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3 years ago

Complete steps to find x +( 7×+24)=128

Mathematics

2 answers:

spin [16.1K]3 years ago

8 0

Answer:

8x=104

Step-by-step explanation:

The answer above is the question simplified down

natita [175]3 years ago

6 0

X = 13 you welcome :)

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Evaluate the following a) 4!= b) 0!= c) 2! + 3!= d) 3! x 4! e) 10! / 7!

IRISSAK [1]

Answer:

See below for answers and explanations

Step-by-step explanation:

4! = 4*3*2*1 = 24

0! = 1

2! + 3! = 2*1 + 3*2*1 = 2 + 6 = 8

3! * 4! = 3*2*1 * 4*3*2*1 = 6 * 24 = 144

10! / 7! = 10*9*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 = 10*9*8 / 1 = 720

6 0

2 years ago

Find the length of the radius of a sphere with a surface area of 5541.77 cm2.

ladessa [460]

The area of a sphere: A = 4.π.R²

A= 5541.77 → 5541.77 = 4.π.R²

R² = (5541.77)/(4.π)
R² = 441; → R =√441; R = 21 cm

6 0

3 years ago

A train is traveling at a constant speed of 105 miles per hour. How many feet does it travel in 3 seconds? Remember that 1

ipn [44]

Answer:

in 3 seconds the train should travel 462 feet

Step-by-step explanation:

So, first, we convert the miles into feet.

105 x 5280 = 554400

This means the train travels 554400 feet per hour.

then, we convert the hour, into second.

Since an hour has 60 minutes, and a minute has 60 seconds, we simply multiply:

1(hour) x 60(minutes) x 60(seconds) = 3600

So the train moves 554400 feet every 3600 seconds.

To get how many feet it travels in one second, we divide this by 3600.

3600/3600 and 554400/3600

1 and 154.

For every second the train travels 154 feet.

Then we simply multiply them by 3.

3x1 and 3x154, which equals:

3 and 462.

So the train travels 462 feet every 3 seconds.

I tried to explain this as simply as I can.

I hope its all correct.

6 0

3 years ago

These two trapezoids are similar What is the correct way to complete the similarity statement?

pentagon [3]

Option A:

$\mathrm{ABCD} \sim \mathrm{GFHE}$

Solution:

ABCD and EGFH are two trapezoids.

To determine the correct way to tell the two trapezoids are similar.

Option A: $\mathrm{ABCD} \sim \mathrm{GFHE}$

AB = GF (side)

BC = FH (side)

CD = HE (side)

DA = EG (side)

So, $\mathrm{ABCD} \sim \mathrm{GFHE}$ is the correct way to complete the statement.

Option B: $\mathrm{ABCD} \sim \mathrm{EGFH}$

In the given image length of AB ≠ EG.

So, $\mathrm{ABCD} \sim \mathrm{EGFH}$ is the not the correct way to complete the statement.

Option C: $\mathrm{ABCD} \sim \mathrm{FHEG}$

In the given image length of AB ≠ FH.

So, $\mathrm{ABCD} \sim \mathrm{FHEG}$ is the not the correct way to complete the statement.

Option D: $\mathrm{ABCD} \sim \mathrm{HEGF}$

In the given image length of AB ≠ HE.

So, $\mathrm{ABCD} \sim \mathrm{HEGF}$ is the not the correct way to complete the statement.

Hence, $\mathrm{ABCD} \sim \mathrm{GFHE}$ is the correct way to complete the statement.

3 0

3 years ago

Describe a time when you use math outside of school ( pls I need help with this writing prompt)

Serga [27]

Answer:

Counting my money

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers